A good number of them share a common factor

We are interested in positive integers $X$ such that $100 \phi(X) \; = \; (100 - M)X$ for some integer $M$ .

One solution is $X=1$ , for which $M=0$ .

If $X > 1$ , and we consider the prime factorisation of $X$ : $X \; = \; p_1^{a_1}p_2^{a_2}\cdots p_N^{a_N}$ where $p_1 < p_2 < ... < p_N$ are primes and $a_1,a_2,...,a_N$ are positive integers, then $\begin{aligned} 100 (p_1-1)(p_2-1)\cdots(p_N-1) p_1^{a_1-1}p_2^{a_2-1} \cdots p_N^{a_N-1} & = (100-M)p_1^{a_1}p_2^{a_2}\cdots p_N^{a_N} \\ 100(p_1-1)(p_2-1)\cdots (p_N-1) & = (100-M)p_1p_2\cdots p_N \end{aligned}$ Since $p_N$ cannot divide $p_1-1,p_2-1,...,p_N-1$ , we deduce that $p_N$ divides $100$ , and so is either $2$ or $5$ .

If $p_N=2$ , then $X = 2^a$ for some $a \ge 1$ , and $M = 50$ .

If $p_N=5$ , then since $3$ does not divide any of $2-1$ , $3-1$ , $5-1$ , $100$ , we see that none of $p_1,p_2,...,p_{N-1}$ can be equal to $3$ .

• We can have $X = 5^b$ for $b \ge 1$ , in which case $M = 20$ .
• We can have $X = 2^a5^b$ for $a,b \ge 1$ in which case $M = 60$ .

Thus the possible values of $X$ are integers of the form $2^a5^b$ for $a,b \ge 0$ , and so $\sum_{n=1}^\infty \frac{1}{x_n} \; = \; \sum_{a,b \ge 0} 2^{-a} 5^{-b} \; = \; \sum_{a=0}^\infty 2^{-a} \sum_{b=0}^\infty 5^{-b} \; = \; 2 \times \tfrac54 \; = \; \boxed{\tfrac52}$